BOTANY

SECTION A

1.

The correct sequence of phases in cell cycle is:

1. G1SG2MG1SG2M

2. MG1G2SMG1G2S

3. G1G2SMG1G2SM

4. SG1G2MSG1G2M

Ans. 1
Sol.

Page 163, XI NCERT


2.

During what phase in the cell cycle would you find the most DNA per cell?
1. G1
2. G2
3. S
4. Prophase II

Ans. 2
Sol.

Page 163, XI NCERT


3.

The figurees below shows 3 phases of mitosis select the option given correct identification together with the correct event ?

1. C- Telophase-Nuclear envelope assembles around the chromosome clusters

2. B- Anaphase-Segregation of homologous chromosomes.

3. A- Prophase-Chromosomes get fully condensed.

4. C- Metaphase-Condensation of chromatin to form chromosome 

Ans. 1
Sol.

Fig.10.2 c to e ;Page 166, XI NCERT


4.

Major event that occurs during anaphase of mitosis which brings about equal distribution of chromosomes is:

1. Condensation of chromatin

2. Replication of genetic material

3. Splitting of centromere

4. Pairing of homologous chromosomes

Ans. 3
Sol.

Page 165-166, XI NCERT


5.

Chromatin condensation and movement of duplicated centriole towards opposite pole can be observed during -

1. Prophase

2. Metaphase

3. Anaphase

4. Telophase

Ans. 1
Sol.

Page 164, XI NCERT


6.

Following are the events occurs during meiosis :

(A) Appearance of chiasmata

(B) Synapsis

(C) Assembly of meiotic spindle

(D) Use of recombinase enzyme

Choose the correct sequence :-

1. ABCD

2. BDAC

3. DCBA

4. BCAD

Ans. 2
Sol.

Page 168, XI NCERT


7.

The microtubules from the opposite poles of the spindle attach to the pair of homologous chromosomes in

1. Metaphase - I 

2. Prophase - I

3. Metaphase 

4. Metaphase - II

Ans. 1
Sol.

Page 168, XI NCERT


8.

Each pole receives half the chromosome number of the
parent cell, is true for which stage?
1. Anaphase - II

2. Anaphase - I

3. Telophase-I 

4. Telophase-II

Ans. 2
Sol.

Page 171, Summary, XI NCERT


9.

A reduction step during meiosis is important because:

(1) It returns the chromosome number to normal before fertilization

(2) There is a mechanism for this

(3) Only one copy of each chromosome is necessary

(4) Otherwise chromosome copies would double each fertilization

Ans. 4
Sol.

Meiosis ensures the production of haploid phase in the life cycle of sexually reproducing organisms whereas fertilisation restores the diploid phase. We come across meiosis during gametogenesis in plants and animals. This leads to the formation of haploid gametes


10.

When does Prophase II start usually?

(1) Chromosomes are fully elongated

(2) Before chromosomes are fully condensed

(3) Before chromosomes are fully elongated

(4) After chromosomes are fully elongated

Ans. 3
Sol.

PAGE 169, XI NCERT


11.

What do A, B, C and D represent in the following figure?

(1) A : carrier protein, B : symport, C : uniport, D : antiport

(2) A : carrier protein, B : uniport, C : antiport, D : symport

(3) A : carrier protein, B : antiport, C : symport, D : uniport

(4) A : carrier protein, B : uniport, C : symport, D : antiport

Ans. 2
Sol.

Page-177, XI NCERT


12.

Apply concept of water potential and osmosis

Solution of which chamber has a lower solute potential?

(1) Chamber A

(2) Chamber B

(3) Both will have equal values

(4) Cannot be predicted

Ans. 2
Sol.

XI NCERT Page-180


13.

Select the correct statement:-

1. Pue water has minimum ψωψω

2. Pure water has maximum ψωψω

3. Pure water has maximum DPD

4. Pure water has variable DPD

Ans. 2
Sol.

Page 179, XI NCERT


14.

Movement of molecule across a typical plant cell i.e. 50 micrometer takes

(1) 10 seconds

(2) 15 seconds

(3) 2.5 seconds

(4) 16 seconds

Ans. 3
Sol.

Page number-183, XI NCERT


15.

Consider the following statements:

I. The apoplast system is continuous throughout the plant, except at the casparian strips of the endodermis.


II. Transport proteins of endodermal cells are control points, where a plant adjusts the quantity and type of solute that reaches the xylem.

III.A C4C4 plant loses about twice as much water as a C3C3 plant for the same amount of CO2CO2fixed.

Which of the above statements are true?

1. I and II only

2. I and III only

3. II and III only

4. I, II and III

Ans. 1
Sol.

Page 184, 189 XI NCERT


16.

Transpiration 

1. Loss of water in soil 

2. Loss of minerals in soil 

3. Loss of water to the air from the leaves 

4. Loss of organic substances because of photorespiration 

Ans. 3
Sol.

Para-3, Page-187, XI NCERT


17.

The process responsible for facilitating loss of water in liquid form from the tip of grass blades at night and early morning is:

1. Root pressure

2. Imbibition

3. Plasmolysis

4. Transpiration

Ans. 1
Sol.

Page 186, XI NCERT


18.

Consider the following two statements:

I. The direction of movement in the phloem is bi-directional.

II. The source-sink relationship in plants is variable.

1. Both I and II are correct and II explains I

2. Both I and II are correct but II does not explain I

3. I is correct but II is incorrect

4. Both I and II are incorrect 

Ans. 1
Sol.

page 190, XI NCERT


19.

Prehistoric man used which of the following methods to split rocks and boulders?

(1) Plasmolysis

(2) Active Transport

(3) Osmosis

(4) Imbibition

Ans. 4
Sol.

Page number-183, XI NCERT


20.

ln above diagram A is having :-

(1) Hypotonic solution 

(2) Hypertonic solution 

(3) Low ψψw

(4) Both B & C correct

Ans. 4
Sol.

Page 181-182, XI NCERT


21.

Alcohol Dehydrogenase needs which of the following ion for activation?

(1) Zn++

(2) Mo

(3) Cr

(4) Mg++

Ans. 1
Sol.

Page number-196, XI NCERT


22.

In the given diagram of a typical set up for nutrient solution culture, the purpose of X is:

1. Addition of water
2. Addition of nutrients
3. Removal of wastes
4. Aeration

Ans. 4
Sol.

Page 195 XI NCERT


23.

Deficiency symptoms of an element tend to appear first in young leaves. It indicates that the element is relatively immobile. Which one of the following elemental deficiency would show such symptoms?

1.  Sulphur

2.  Magnesium

3.  Nitrogen

4.  Potassium

Ans. 1
Sol.

Page 198, XI NCERT


24.

Which of the following is not associated with potassium ions?

(1) Maintain anion-cation balance in cells

(2) Involved in protein synthesis

(3) Involved in opening and closing of stomata and activation of enzyme

(4) Maintenance of the turgidity of the cells and all phosphorespiration reactions

Ans. 4
Sol.

Page number-197, XI NCERT


25.

If by radiation all nitrogenase enzyme are inactivated, then there will be no :-

(1) Fixation of atmospheric nitrogen

(2) Conversion from nitrate to nitrite in legumes

(3) Conversion from ammonium to nitrate in soil

(4) Fixation of nitrogen in legumes

Ans. 4
Sol.

Page 202, XI NCERT


26.

Which of the following is an amide involved in nitrogen metabolism by plants?

1. Methionine

2. Cysteine

3. Serine

4. Asparagine

Ans. 4
Sol.

Page 204, XI NCERT


27.

(a) Soil are reservoir of all mineral elements that are essential for the proper growth and development of plants.

(b) Initial uptake of minerals into the symplast is slow

(c) Uptake of minerals in inner space is rapid

1. All are correct

2. Only (a) is incorrect

3. (b) & (c) are incorrect

4. Only (c) is incorrect

Ans. 4
Sol.

Page 200, XI NCERT


28.

The main amino acid over which Transamination occurs.

(1) Glutamate

(2) Alanine

(3) Glycine

(4) Lysine

Ans. 1
Sol.

Page number-204 12.6 Metabolism of Nitrogen


29.

Plants obtain Iron in the form of

(1) Ferric ions

(2) Ferrous ions

(3) FeO

(4) Both B and C

Ans. 1
Sol.

Page number-197, XI NCERT


30.

Which of the following statement is correct?
a. H2S is the hydrogen donor for purple and green sulphur bacteria
b. H2O is the hydrogen donor for purple and green sulphur bacteria
c. Oxygen is evolved by the purple and green sulphur bacteria
d. Oxygen evolved by the green plants comes from the CO2.

Ans. 1
Sol.

Page 208 XI NCERT


31.

In the given graph what does A, B represent?

       

1. A - Absorption spectrum; B - Action spectrum (Chl. a).

2. A - Action spectrum; B - Absorption spectrum (carotenoids).

3. A - Absorption spectrum; B - Action spectrum (Ch. b).

4. A - Action spectrum; B - Absorption spectrum (Ch.a).

Ans. 4
Sol.

Page 210, XI NCERT


32.

Cyclic photophosphorylation is different from non-cyclic photophosphorylation as the former

1. Is performed by a collaboration of both PS-II and PS-I

2. Does not require any external source of electrons

3. Is connected with photolysis of water

4. Is connected with ATP and NADPH production

Ans. 2
Sol.

Page 213, XI NCERT


33.

The synthesis of one molecule of glucose during the Calvin cycle requires.

1. 12 molecules of ATP and 18 molecules of NADPH2NADPH2

2. 6 molecules of ATP and 12 molecules of NADPH2NADPH2

3. 18 molecules of ATP and 12 molecules of NADPH2NADPH2

4. 12 molecules of each of ATP and NADPH2NADPH2

Ans. 3
Sol.

Page 218, XI NCERT


34.

In C44 plants CO22 are reduced into sugar is : 

1. Mesophyll cells

2. In bundle sheath cells

3. In spongy parenchyma

4. Both A & B

Ans. 2
Sol.

Page 219, XI NCERT


35.

A plant in your garden avoids photorespiratory losses, has improved water use efficiency, shows high rates of photosynthesis at high temperatures and has improved efficiency of nitrogen utilization. In which of the following physiological groups would you assign this plant?

(1) C4

(2) CAM

(3) Nitrogen-fixer

(4) C3

Ans. 1
Sol.

Page 218, XI NCERT


SECTION B

36.

Identify the symbols A, B and C in the figure given below

 

        A

         B

        C

(1)

G0

Prophase

Cytokinesis

(2)

Prophase

Metaphase

Telophase

(3)

G1

S

G2

(4)

Cytokinesis

Prophase

G0

1. (1)

2. (2)

3. (3)

4. (4)

Ans. 4
Sol.

Page 163, XI NCERT


37.

A stage in cell division is shown in the figure. Select the answer which gives correct

identification of the stage with its characteristics.

                             

1. Telophase       - Nuclear envelope reforms, Golgi complex reforms

2. Late anaphase -  Chromosomes move away from equatorial plate, Golgi complex not

present

3. Cytokinesis    -  Cell plate formed, mitochondria distributed between two daughter cells

4. Telophase      -  Endoplasmic reticulum and nucleolus not reformed yet

Ans. 1
Sol.

PAGE 166, XI NCERT


38.

“The synaptonemal complex is formed during _A_
stage and dissolves during _B_ stage”.
Complete the above statement by choosing the correct
option for A and B
         A               B
1. Diplotene    Diakinesis
2. Leptotene   Zygotene
3. Zygotene    Diplotene
4. Pachytene  Diplotene

Ans. 3
Sol.

Page 168, XI NCERT


39.

In the diagram given below showing the pathway of water movement in the root, A, B, C and D respectively represent:

1. Symplastic pathway, Apoplastic pathway, Phloem and Xylem
2. Apoplastic pathway, Symplastic pathway, Phloem and Xylem
3. Symplastic pathway, Apoplastic pathway, Xylem and Phloem
4. Apoplastic pathway, Symplastic pathway, Xylem and Phloem

Ans. 3
Sol.

Page 185, XI NCERT


40.

Which of the following facilitates opening of stomatal aperture?

1. Decrease in turgidity of guard cells

2. Radial orientation of cellulose microfibrils in the cell wall of guard cells.

3. Longitudinal orientation of cellulose microfibrils in the cell wall of guard cells

4. Contraction of outer wall of guard cells

Ans. 2
Sol.

PAGE 187, XI NCERT


41.

The incorrect statement for facilitated diffusion is
1. Special proteins help move substances across
membranes without expenditure of energy.

2. Can cause net transport of molecules from a low
to a high concentration

3. It is very specific, it allows cell to select substances
for uptake.

4. Transport rate reaches a maximum when all of
the protein transporters are being used

Ans. 2
Sol.

Page 176, XI NCERT


42.

 A mature corn plant absorbs…….. of water in a day while a mustard plant absorbs water equal to its own weight in about…….. 

1. 5L, 3hours 

2. 3L, 5 hours 

3. 5L, 5 hours 

4. 3L, 3 hours 

Ans. 2
Sol.

Page number-179, XI NCERT


43.

The criteria for essentiality of an element for a plant include all except:
1. It must be absolutely necessary for supporting normal growth and reproduction
2. The requirement must be specific and not replaceable by another element
3. It must be directly involved in the metabolism of the plant.
4. They must be present in the plants in a concentration in excess of 10 mmole/Kg of dry matter.

Ans. 4
Sol.

Page 195, XI NCERT


44.

Julius Von Sachs
Which of the following is not related?

1. A prominent German Botanist
2. In 1860, he demonstrated for the first time that plants could be grown to maturity in a defined nutrient solution in complete absence of soil
3. The technique, Hydroponics
4. Categorized the essential and non Essential mineral nutrition.

Ans. 4
Sol.

Page number-194, XI NCERT


45.

The deficiency symptoms of mobilizable substances appear first in Older parts because

(1) In order to mobilize the elements to young part, the biomolecules of older parts are broken down

(2) In order to mobilize the elements the older part becomes weaker

(3) The work of mobilization is random and takes a lot of energy to cause it

(4) All of these

Ans. 1
Sol.

Page number-198, XI NCERT


46.

Nitrification does not include

(1) Conversion of ammonia to nitrite

(2) Conversion of nitrite to nitrate

(3) Conversion of nitrogen to ammonia

(4) Both B and C

Ans. 3
Sol.

Page number-201, XI NCERT


47.

Match the following concerning essential elements and their functions in plants :

(a) Iron            (i) Photolysis of water

(b) Zinc           (ii) Pollen germination

(c) Boron         (iii) Required for chlorophyll biosynthesis

(d) Manganese  (iv) IAA biosynthesis

Select the correct option:

     (a)  (b)  (c)  (d)

1.  (iv) (iii)  (ii)  (i)

2.  (iii)  (iv) (ii)  (i)

3.  (iv)  (i)  (ii)  (iii)

4.  (ii)   (i)  (iv)  (iii)

Ans. 2
Sol.

Page 197-198, XI NCERT


48.

According to the chemiosmotic hypothesis, there is a basic difference between respiration and photosynthesis. This is _____

1. In chloroplast, H+ accumulation takes place in the lumen of thylakoid but in mitochondria, this takes place in perimitochondrial space.

2. Accumulation of H+ in chloroplast takes place in matrix i.e., stroma and in mitochondria, this takes place in cristae.

3. In both the mitochondria and chloroplast the accumulation of H+ takes place in inter-membranous space but with a high rate in mitochondria

4. Both 1 and 3 are correct.

Ans. 1
Sol.

Page 213, XI NCERT


49.

What is incorrect about bundle sheath cells-
a. Multilayered
b. large number of chloroplasts
c. Intercellular spaces
d. Thick walls

Ans. 3
Sol.

Page 218 XI NCERT


50.

How many statements are correct?
i. Products of dark reaction are substrates of light reaction.
ii. Products of light reaction are substrates of dark reaction.
iii. In the stroma, enzymatic reactions incorporate CO2 into the plant leading to the synthesis of sugar.
iv. Dark reactions are directly light-driven.
v. Chloroplast is a single membranous organelle.

1. 2
2. 3
3. 4
4. 1

Ans. 1
Sol.

Page 209, XI NCERT


BOTANY

SECTION A

51.

In mammals, the teeth are.
I. only two sets, present throughout life.
II. embedded in the socket of the jaw bones.
III. different types.

These conditions are respectively referred as
1. diphyodont, heterodont and thecodont
2. diphyodont, thecodont and heterodont
3. the codont, diphyodont and heterodont
4. thecodon, heterodont and diphyodont

Ans. 2
Sol.

PAGE 257 and 258, XI NCERT


52.

Brunner's gland are characteristic feature of:-

(1) jejunum of small intestine

(2) ileum

(3) duodenum

(4) fundic region of stomach

Ans. 3
Sol.

Last paragraphs, PAGE 259 and 262, XI NCERT


53.

Mark the incorrect match regarding the process involved
in digestion and absorption?
1. Muscular activities              Hormonal control.
of different parts of the
alimentary canal.
2. Absorption and                  Passive process and
transport of water.                       depends upon the
                                                  osmotic gradient.
3. Absorption of fatty             Involvement of micelle.
acid.
4. Large intestine.               Absorption of some water, minerals and certain drugs.

Ans. 1
Sol.

Page 264, XI NCERT


54.

In the given diagram showing anatomical regions of the human stomach, A,B and C respectively represent :

1. Fundus, Cardia and Pylorus

2. Cardia, Pylorus and Fundus

3. Fundus, Pylorus and Cardia

4. Pylorus, Cardia and Fundus

Ans. 1
Sol.

Page 259, XI NCERT


55.

Which of the following structures prevents the back 

flow of faecal matter?

(1) Cardiac sphincter

(2) Pyloric sphincter

(3) lleo-caecal value

(4) Sphincter of Oddi

Ans. 3
Sol.

Page 264, XI NCERT


56.

Which of the following is true about digestion and 

absorption of food?

(1) Trypsin, chymotrypsin and carboxypeptidase are brush border enzymes

(2) The chemical process of digestion is initiated in the oral cavity. About 60% Starch is hydrolysed here by salivary amylase to form maltose

(3) Rennin is a proteolytic enzyme found in gastric juice of infants. Small amount of lipases are also secreted by the gastric glands

(4) If breast feeding is replaced by less nutritive food low in proteins and calories, the infants below the age of one year will suffer from Kwashiorkar

Ans. 3
Sol.

Page 262, XI NCERT


57.

In jaundice - what is not observed?

(1) Liver is affected

(2) Skin and eyes turn yellow because of deposition of bile salts

(3) Spreads due to contaminated water

(4) More than one option is correct.

Ans. 2
Sol.

NCERT page number 265.


58.

Match the items given in Column - I with those in Column - II and choose the correct option.

 

Column-I

 

Column-II

(a)

Rennin

(i)

Vitamin B12

(b)

Enterokinase

(ii)

Facilitated transport

(c)

Oxyntic cells

(iii)

Milk proteins

(d)

Fructose

(iv)

Trypsinogen

1. (a) - (iii), (b)- (iv), (c)- (ii), (d)-(i)

2. (a) - (iv), (b)- (iii), (c)- (i), (d)-(ii)

3. (a) - (iv), (b)- (iii), (c)- (ii), (d)-(i)

4. (a) - (iii), (b)- (iv), (c)- (i), (d)-(ii)

Ans. 4
Sol.

Page 262, XI NCERT


59.

Which of the following enzyme is not included in

the secretions of the brush border cells of 

mucosa?

(1) Maltase

(2) Dipetidase

(3) Nucleosidease

(4) Carboxypeptidase

Ans. 4
Sol.

Page 262, XI NCERT


60.

When breast feeding is replaced by less nutritive food low in proteins and calories; the infants below the age of one year are likely to suffer from 

1. marasmus

2. rickets

3. kwashiorkor

4. pellagra

Ans. 1
Sol.

PAGE 266, XI NCERT


61.

Match the column I with column II :-
    Column-I                       Column-II
(i) Vomiting                  (a) Inadequate enzyme secretion
(ii) Diarrhoea                (b) Irregular bowel movement
(iii) Constipation           (c) Increased liquidity of faecal discharge
(iv) Indigestion             (d) A feeling of nausea

1. (i)-a, (ii)-c, (iii)-b, (iv)-d
2. (i)-d, (ii)-b, (iii)-c, (iv)-a
3. (i)-d, (ii)-c, (iii)-b, (iv)-a
4. (i)-a, (ii)-b, (iii)-c, (iv)-d

Ans. 3
Sol.

Page 265, XI NCERT


62.

Carrier proteins faciliate the absorption of substance like

1. amino acids and glucose

2. glucose and fatty acids

3. fatty acids and glycerol

4. fructose and some amino acids

Ans. 1
Sol.

PAGE 264, XI NCERT


63.

Which of the following is true about epiglottis
1. thin elastic cartilaginous flap
2. prevent food entry into larynx
3. Coughing occurs due to its improper movement
4. All of these

Ans. 4
Sol.

NCERT, Heading 17.1.1, para 1, page-269.


64.

Choose the wrong statement
(1)CO2 is the harmful molecule and released during catabolic reactions
(2)respiration will always require O2
(3)Breathing is a part of respiration
(4)All of the above

Ans. 2
Sol.

NCERT, Introduction, para 1, page 268.


65.

By the contration in radial muscle of diaphragm it become flattened in shape, so, volume of thoracic cavity increases in :-

1. Anterior-posterior axis

2. Dorsal-ventral axis

3. Anterior-lateral axis

4. Dorsal-lateral axis

Ans. 1
Sol.

Page 270, XI NCERT


66.

The amount of air that enters the lungs during normal, restful breathing is called the______.

1. Vital capacity

2. Tidal volume

3. Total lung capacity

4. Expiratory reserve volume 

Ans. 2
Sol.

Page 271, XI NCERT


67.

Total volume of air a person can expire after a normal inspiration is 

A. Tidal volume               B. TV + ERV

C. Expiratory capacity      D. ERV + RV

1. A only

2. B & C only

3. A, B & C only

4. A,B,C & D

Ans. 2
Sol.

Page 272, XI NCERT


68.

Diffusion membrane consists of

1. Squamous epithelium of alveoli

2. endothelium of alveolar capillaries

3. Basement substance between them

4. All of the above

Ans. 4
Sol.

NCERT, Heading 17.3, page-273


69.

Given below a diagram of a section of an alveolus
with a pulmonary capillary
Which of the following is a correct statement for
diffusion of gases?

1. Diffusion of O2 and CO2
from A to B or B to A takes place with the same rate.

2. O2 will diffuse faster from A to B than CO2
from B to A

3. Only O2 will diffuse from A to B not CO2
from B to A
4. Only CO2 will diffuse from B to A, not O2
from A to B

Ans. 2
Sol.

Page 273, XI NCERT


70.

Which of the following statements is correct regarding CO2 and O2CO2 and O2 transport in blood ?

1. 100 ml of oxygenated blood carries 53 ml of O2O2

2. 100 ml of blood always transport equal amounts of O2 and CO2O2 and CO2

3. 100 ml of oxygenated blood delivers more O2O2 than CO2CO2 to the tissues

4. O2 and CO2O2 and CO2 are both maximally transported through the blood plasma 

Ans. 3
Sol.

Page 274-275, XI NCERT


71.

Which of the following match is correct?

1. Emphysema : reduction of surface area of alveoli and bronchi

2. Pneumonia : occupational disease with asbestos

3. Silicosis : inflammation of alveoli

4. Asthma : excessive secretion of bronchial mucus

Ans. 1
Sol.

PAGE 275, XI NCERT


72.

Every 2000 mL of deoxygenated blood delivers approximately, how much mL CO22 to alveoli?

1. 4 mL

2. 4.2 mL

3. 8 mL

4. 80 mL

Ans. 4
Sol.

Page 275, XI NCERT


73.

Bulk of carbon dioxide (CO2) released from
body tissues into the blood is present as :
1. carbamino-haemoglobin in RBCs
2. bicarbonate in blood plasma and RBCs
3. Free CO2 in blood plasma
4. 70% carbamino-haemoglobin and 30% as
bicarbonate

Ans. 2
Sol.

Page 274, XI NCERT


74.

A center that moderates the functions of the respiratory rhythm center is located in:

1. Dorsal medulla oblongata

2. Ventral medulla oblongata

3. Pons Varolii

4. Pre central gyrus of the cerebrum

Ans. 3
Sol.

Read page 275, Para-2, XI NCERT.


75.

Which of the following cell will play a role in resistance of infection and allergic reaction?

(1) Eosinophil

(2) Basophil

(3) Lymphocyte

(4) Neutrophils

Ans. 1
Sol.

PAGE 279-280, XI NCERT


76.

Choose the incorrect match w.r.t blood components

1. Leucocytes – 6000 - 8000 mm–3

2. Globulins – Involved in defence mechanisms

3. O blood group – Antibodies A and B absent

4. Thrombokinase – Converts prothrombin into thrombin

Ans. 3
Sol.

Table 18.1 Page 280, XI NCERT


77.

Erythroblastosis fetalis can be prevented during a second pregnancy in a Rh negative mother who is likely to carry a Rh positive fetus by:

1. Administering anti – Rh antibodies to the mother just before the delivery of the second child

2. Administering Rh antigen to the mother just after the delivery of the first child

3. Administering Rh antigen to the mother just before the delivery of the second child

4. Administering anti – Rh antibodies to the mother just after the delivery of the first child

Ans. 4
Sol.

Page 281, XI NCERT


78.

Fibrins are formed by conversion of _______ in _______ by the enzyme _______

(1) Fibrinogen, serum, Thrombokinase

(2) prothrombin, plasma, thrombin

(3) Fibrinogen, plasma, thrombin

(4) Prothrombin, blood, thrombokinase

Ans. 3
Sol.

PAGE 281, XI NCERT


79.

The inter-atrial septum in the human heart can be best described as:

1. A thin muscular wall

2. A thick muscular wall

3. A thin fibrous tissue

4. A thick fibrous tissue

Ans. 1
Sol.

Read page 283, XI NCERT


80.

The passage of blood from left atrium to body is

(1) left atrium – left ventricle – Aorta – Body

(2) left atrium – right ventricle – Aorta – Body

(3) left atrium – left ventricle – Pulmonary Artery – Body

(4) left atrium – left ventricle – Pulmonary Veins – Body

Ans. 1
Sol.

XI NCERT, fig.18.4. P -286-287, Double Circulation


81.

The pumping pressure of healthy heart is

(1) 120 mmHg

(2) 80 mmHg

(3) 140 mmHg

(4) 90 mmHg

Ans. 1
Sol.

(1) XI NCERT, P-287 Disorders of Circulatory System


82.

Which of the following sequences best represents the pathway of an action potential through the Heart's conduction system?

(i) Sino-atrial (SA) node

(ii) Purkinje fibres

(iii) Bundle of His

(iv) Atrio-ventricular (AV) node

(v) Right and left bundle branches

1. (i),(iv),(iii),(ii),(v)

2. (iv), (i),(iii),(v),(ii)

3. (iii),(iv),(i),(ii),(v)

4. (i),(iv),(iii),(v),(ii)

Ans. 4
Sol.

Page 284, XI NCERT


83.

In ECG the ventricular contraction occurs

1.  Just after P wave and before Q wave.

2.  Just after Q wave but before T wave.

3.  In between S-T segment.

4.  After the end of T wave.

Ans. 2
Sol.

PAGE 286, XI NCERT


84.

Consider the following four statements (a-d) and select the option which includes all the correct :-

(a) Coronary Artery disease, (CAD) often referred
to as Atherosclerosis

(b) Heart failure means when the heart muscle is
suddenly damaged by an inadequate blood supply

(c) High blood pressure leads to heart diseases and
also affects vital organs like brain and kidney

(d) Angina occurs due to conditions that affect the
blood flow

Options :

1. Statements (b), (c) and (d).
2. Statements (a), (b).
3. Statements (b), (d).
4. Statement (a), (c) and (d).

Ans. 4
Sol.

Page 288, XI NCERT


85.

Read the following statements:-

(1) Lymph is devoid of RBC
(2) Exchange of nutrients, gasses etc, between blood and the tissue cell always occur through the lymph
(3) Tissue fluid is also called as lymph.
(4) Lymph is a colourless fluid containing all types of WBCs which are responsible for the immune response of the body.

How many statements are correct?

1. 2

2. 3

3. 4

4. 1

Ans. 2
Sol.

Page 282, XI NCERT


SECTION B

86.

Which of the following are not correct:

(i) Spleen is the graveyard for RBC

(ii) All the mammals have biconcave RBCs

(iii) Clotting factors present in plasma are in active form.

(iv) All the substances released by platelets are involved in clotting only.

(1) (i), (iii), (iv)

(2) (ii), (iii), (iv)

(3) (i), (ii), (iii), (iv)

(4) (ii), (iii)

Ans. 2
Sol.

NCERT, P -279, Formed Elements, Para-1


87.

Which of the following reactions is not catalysed

by a brush border enzyme?

(1) Maltose MaltaseMaltase gulcose+ gulcose

(2) Lactose LactaseLactase gulcose + gulcose

(3) Starch AmylaseAmylase Disaccharide

(4) Nucleotides NucleotidaseNucleotidase Nucleoside

 

Ans. 3
Sol.

Page 263, XI NCERT


88.

Which of the following do not belong to category of brush border enzymes?

1. Lactase, nucleotidase

2. Aminopeptidase, sucrase

3. Procarboxypeptidase, steapsin

4. Nucleosidase, maltase

Ans. 3
Sol.

Page 262, XI NCERT


89.

Which of the following is not the function of the large intestine?

1.  Absorption of some water, minerals and certain drugs

2.  Nutrient absorption

3.  Secretion of mucus to lubricate feces

4.  The temporary storage of feces in rectum

Ans. 2
Sol.

Page 264, XI NCERT


90.

How many of the following statements are wrong?

a. The bile duct and the pancreatic duct open 

      together into the duodenum as the common

      hepato-pancreatic duct which is guarded by 

      sphincter of Boyden.

b. The parotid, sub-maxillary and sub-lingual

      salivary glands are located just outside the 

      buccal cavity.

c. In duodenum, glands are only found in 

     submucoas which are Brunner's glads, a 

     type of simple branched tubular glands.

d. There is no modification of the four layers

      namely serosa, muscularis, sub-mucosa and

      mucosa in differnt parts of alimentary canal.

e. lipases are absent in secreation of gastric

    glands.

(1) One

(2) Two

(3) Three

(4) Four

Ans. 4
Sol.

Page 259-262, XI NCERT


91.

Which of the following is not observed in Diarrhoea?

(1) Abnormal frequency of bowl movement

(2) Increased liquidity of faecal discharge

(3) Reduction of Absorption of food

(4) Reduction of digestion of food

Ans. 4
Sol.

NCERT page number 265


92.

In the given figure, identify the muscles involved in raising the rib cage

1. External intercostals 

2. Internal intercostals

3. External obliques

4. Abdominal recti

Ans. 1
Sol.

Page 270-271, XI NCERT


93.

The partial pressure of O2O2 and CO2CO2 in the arterial blood respectively is

1. 40 mmHg and 45 mmHg

2. 100 mmHg and 45 mmHg

3. 95 mmHg and 40 mmHg

4. 45 mmHg and 40 mmHg

Ans. 3
Sol.

Table 17.1 and Fig.17.2 Page 272-273, XI NCERT


94.

Which of the following statements are correct w.r.t. transportation of gases?

A. Blood transport CO2CO2 comparatively easily because of its higher solubility

B. CO2CO2 diffuses into blood, passes into RBCs and reacts with water to form H2CO3H2CO3

C. 20-25 percent CO2CO2 is carried by haemoglobin as carbamino-haemoglobin

D. Oxyhaemoglobin of erythrocytes is basic in nature and it increases pH of blood

1. A only

2. A & B only

3. A, B & C only

4. A, B, C & D

Ans. 3
Sol.

Page 273-275, XI NCERT


95.

Read the following four statements (a-d) and select
the option which includes all correct ones only :-
(a) Exchange of O2 and CO2 at alveoli and tissue
occur by active transport.
(b) Long exposure to industrial dust leads to
inflammation leading to fibrosis and thus
causing serious lung damage.
(c) EICM and IICM are muscles actively involved in
normal and forced breathing respectively.
(d) Spirometer is unable to find out the functional
residual capacity and total lung capacity.

1. b, c and d
2. b and d
3. a, b and d
4. a, b, c and d

Ans. 1
Sol.

Page 270- 272 and 276, XI NCERT


95.

Choose the incorrect statement

1. Every 100 ml of deoxygenated blood delivers approximately 4 ml of CO2 to the alveoli

2. Carbonic anhydrase is present in very high concentration in RBC

3. High pCO2 and low pO2 in tissues help in binding of carbon dioxide

4. CO2 is carried in haemoglobin as carboxyhaemoglobin 

Ans. 4
Sol.

Page 274-275, XI NCERT


96.

Which of the following does not shift the oxy-haemoglobin dissociation curve to the right?

1. increased pH

2. increased carbon dioxide

3. increased temperature

4. increased 2,3 -DPG

Ans. 1
Sol.

PAGE 274, XI NCERT


97.

Choose the incorrect match
(1) Cardiac arrest – Condition in which heart

                                  stops beating

(2) T-wave –          Represents depolarization

                            of ventricles

(3) SAN –             Generates maximum
                           number of action
                           potentials

(4) P-wave –          Represents contraction of

                          both atria

Ans. 2
Sol.

Page 286, XI NCERT


98.

What is the effect of parasympathetic neural signals on Heart Activity?

(1) decreases the rate of heart beat

(2) decreases the speed of conduction of action potential

(3) decreases the cardiac output

(4) all of these

Ans. 4
Sol.

(4) XI NCERT, P-287 Regulation of Cardiac Activity


99.

In a mm3 of blood, the maximum number and minimum number of cells would be respectively

1. WBCs, RBCs

2. RBCs, WBCs

3. RBCs, Platelets

4. Platelets, WBCs

Ans. 2
Sol.

Heading 18.1.2, Page 279, 280, XI NCERT


100.

Serum is-

(1) Blood - corpuscles

(2) Blood - corpuscles - clotting factors

(3) Plasma - clotting factors

(4) Both (2) and (3)

Ans. 2
Sol.

Page 282, XI NCERT


CHEMISTRY

SECTION A

151.

The number of σ and π bonds in the molecule C6H6C6H6 are -

1. 6 C – C sigma (σC-C)σCC) bonds, 5 C–H sigma ((σC-H)(σCH) bonds, and 3 C=C pi (πC-CπCC

2. 6 C – C sigma (σC-C)σCC) bonds, 5 C–H sigma ((σC-H)(σCH) bonds, and 2 C=C pi (πC-CπCC

3. 6 C – C sigma (σC-C)σCC) bonds, 6 C–H sigma ((σC-H)(σCH) bonds, and 3 C=C pi (πC-CπCC

4. 6 C – C sigma (σC-C)σCC) bonds, 6 C–H sigma ((σC-H)(σCH) bonds, and 2 C=C pi (πC-CπCC

 

 

Ans. 3
Sol.

152.

The IUPAC name of the above-mentioned compound is - 

1. cis-2-Chloro-3-iodo-2-pentene
2. trans-2-Chloro-3-iodo-2-pentene
3. cis-3-Iodo-4-chloro-3-pentene
4. trans-3-Iodo-4-chloro-3-pentene

Ans. 2
Sol.

153.

The pair of structures that does not represent isomers is -

1.   

2.  

3.  

4.  

Ans. 4
Sol.

154.

 The structure of diphenylmethane is given below:

 

The number of  structural isomers possible when one of the hydrogen atom is replaced by a chlorine atom is -

1. 6       

2. 4       

3. 8       

4. 7

Ans. 2
Sol.

155.

The most stable carbocation among the following is-
                    +1. CH33C CH2                    +1. (CH3)3C CH2                       

2. (CH3)3C+2. (CH3)3C+

                   +3. CH3CH2CH2                   +3. CH3CH2CH2                                    

 4.CH3 C+ H CH2CH34.CH3 C+ H CH2CH3

Ans. 2
Sol.

156.

The correct order of acidity among the following is-

1. CH2=CH2 > CH3-CH=CH2 > CH3C≡CH > CH≡CH

2. CH≡CH > CH3-C≡CH > CH2=CH2 > CH3 CH3

3. CH≡CH > CH2=CH2 > CH3-C≡CH > CH3-CH3

4. CH3-CH3 > CH2=CH2 > CH3-C≡CH > CH≡CH

Ans. 2
Sol.

157.

C|HH2-C|ClH2alk.NaOHCH2=CH2C|HH2C|ClH2alk.NaOHCH2=CH2

Most probable mechanism for this reaction is-

1. E1                                         

2. E2

3. E1CBE1CB                                     

4. αα elimination

Ans. 2
Sol.

158.

An organic compound contains 69% carbon, and 4.8% hydrogen, the remainder being
oxygen. The masses of carbon dioxide, and water produced when 0.20 g of this substance is subjected to complete combustion would be respectively -

1. 0.506 g , 0.0864 g

2. 0.906 g , 0.0864 g

3. 0.0506 g , 0.864 g

4. 0.0864 g, 0.506 g

Ans. 1
Sol.

159.

Match Column I with Column II.

Column I

Column II

A. Dumas method

1. AgNO3

B. Kjeldahl’s method

2. Silica gel

C. Carius method

3. Nitrogen gel

D. Chromatography

4. Ammonium sulphate

Codes

          A          B          C          D         

1.       3          4           1           2

2.       1          2           3           4

3.       1          4           3           2

4.       4          1           3           2

Ans. 1
Sol.

160.

The correct IUPAC name of the given compound is-

1.  3-Ethyl-4-ethenylheptane

2.  3-Ethyl-4-propylhex-5-ene

3.  3-(1-Ethyl propyl) hex-1-ene

4.  4-Ethyl-3-propylhex-1-ene

Ans. 4
Sol.

161.

The final product in the above sequence of reactions is-

 

Ans. 2
Sol.

162.

The most suitable reagent among the following is used to distinguish compound (III) from the rest of the compounds is -

I. CH3CCCH3CH3CCCH3

II. CH3-CH2-CH2-CH3CH3CH2CH2CH3

III. CH3-CH2-CCHCH3CH2CCH

IV. CH3-CH=CH2CH3CH=CH2

1. Br2Br2/CCl4CCl4

 2. Br2Br2/CH3COOHCH3COOH

3. Alk.KMnO4KMnO4 

4. Ammoniacal AgNO3AgNO3

Ans. 4
Sol.

163.

The correct order of relatives rates of hydrogenation of alkenes is

1. Ethylene > propene > 2-butene > 2-methyl-2-butene

2. 2-methyl-2-butene > 2-butene > Propene > Ethylene

3. 2-butene > propene > ethylene > 2-methyl-2-butene

4. Propene > 2-butene > ethylene > 2-methyl-2-butene

Ans. 1
Sol.

164.

Arrange the following carbanions in order of their decreasing stability.

A. H3C-CC-H3CCC

B. H-CC-HCC

C. H3C-CH-2H3CCH2

1. A>B>C

2. B>A>C

3. C>B>A

4. C>A>B

Ans. 2
Sol.

165.

AlCl3CH3-C||O-ClA80%triethylene glycol, heatNH2-NH2, NaOHB73%AlCl3CH3C||OCl(A)80%triethylene glycol, heatNH2NH2, NaOH(B)73%

The product (B) in the above mentioned reaction is -

1. 

2. 

3. 

4. 

 

 

Ans. 2
Sol.

166.

Nitrobenzene on reaction with conc. HNO3/H2SO4 at 80-100°°C forms -

1. 1,2-Dinitrobenzene

2. 1,3-Dinitrobenzene

3. 1,4-Dinitrobenzene

4. 1,2,4-Trinitrobenzene

Ans. 2
Sol.

167.

Benzene reacts with CH3Cl in the presence of anhydrous AlCl3 to form-

1. Toluene

2. Chlorobenzene

3. Benzylchloride

4. Xylene

Ans. 1
Sol.

168.

The reaction/s involved during the formation of photochemical smog are:

1. NO2(g)NO(g)+O(g)1. NO2(g)NO(g)+O(g)

2.O(g)+O2(g)O3(g)2.O(g)+O2(g)O3(g)

3. O3(g)+NO(g)NO2(g)+O2(g)3. O3(g)+NO(g)NO2(g)+O2(g)

4. All of the above

Ans. 4
Sol.

169.

An incorrect statement regarding acid rain is : 

1. It is formed due to the reaction of SO2, and NO2SO2, and NO2 with rainwater.

2. Causes no damage to monuments like the Taj Mahal.

3. It is harmful to plants.

4. Its pH is less than 5.6

Ans. 2
Sol.

170.

The ozone layer of the upper atmosphere is being destroyed by : 

1. Sulphur dioxide

2. Carbon dioxide

3. Chlorofluorocarbon

4. Smog

Ans. 3
Sol.

171.

Green chemistry means such reactions that-

1. Produce colour during reactions.

2. Reduce the use and production of hazardous chemicals.

3. Are related to the depletion of the ozone layer.

4. Study the reactions in plants.

 

Ans. 2
Sol.

172.

Photochemical smog formed in congested metropolitan cities mainly consists of : 

1. Ozone, SO2, and hydrocarbons

2. Ozone, PAN, and NO2

3. PAN, smoke and SO2

4. Hydrocarbons, SO2, and CO2

Ans. 2
Sol.

173.

A farmer was told that fishes were not fit for human consumption because a large amount of pesticides had accumulated in the tissues of fishes. This happen :

1. By transfer of pesticides in the food chain.

2. By degradation of pesticides.

3. By the reaction of pesticides with atmospheric oxygen.

4. By the reaction of pesticides with atmospheric carbon dioxide.

Ans. 1
Sol.

174.

Match the pollutant (s) in Column I with effect (s) in Column II.

Column I

Column II

A. Oxides of Sulphur

1. Red haze in traffic and congested areas

B. Nitrogen dioxide

2. Damage to kidney

C.  Lead

3. ‘Blue baby’ syndrome

D. Nitrate in drinking water

4. Respiratory diseases

Codes

          A         B          C           D

1.       2          3           4           1

2.       1          2           3           4

3.       1          4           3           2

4.       4          1           2           3

Ans. 4
Sol.

175.

The chemical that is an excellent ‘green’ solvent as well as a greenhouse gas : 

1. Methanol

2. CFCs

3. Carbon monoxide

4. Carbon Dioxide

Ans. 4
Sol.

176.

Assertion: The stability of a crystal is reflected by the magnitude of its melting point.

Reason: Greater the intermolecular forces of attraction, the more will be the stability and melting point of the substance.

1.  Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
2.  Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
3.  Assertion is true, but Reason is false.
4.  Both Assertion and Reason are false.

Ans. 1
Sol.

177.

TiO2 is a well-known example of:

1. Triclinic system.

2. Tetragonal system.

3. Monoclinic system.

4. None of the above.

Ans. 2
Sol.

178.

CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being 6.02×1023 mol-1, the density of CsBr is :

1. 42.5 g/cm3g/cm3

2. 0.425 g/cm3g/cm3

3. 8.25 g/cm3g/cm3

4. 4.25 g/cm3g/cm3

Ans. 4
Sol.

179.

In a compound, atoms of element Y form ccp lattice, and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be :-

1. X3Y4

2. X4Y3 

3. X2Y3

4. X2Y

Ans. 2
Sol.

180.

The density of KCl is 1.9893 g cm-3 and the length of a side unit cell is 6.29082 Å as

determined by X-ray diffraction. The value of Avogadro’s number calculated from these

data is:-

1. 6.017 x 1023

2. 6.023 x 1022

3. 7.03 x 1023

4. 6.01 x 1019

Ans. 1
Sol.

181.

A ferric oxide crystallizes in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. The formula of the ferric oxide is-

1. FeO

2. Fe2O3

3. Fe3O4

4. Fe3O2

Ans. 2
Sol.

182.

A: In hcp structure, each sphere is surrounded by eight other spheres.

R: ABC, ABC ................................ represents hcp structure.

1. Both Assertion & Reason are true and the reason is the correct explanation of the assertion.

2. Both Assertion & Reason are true but the reason is not the correct explanation of the assertion.

3. Assertion is true statement but Reason is false.

4. Both Assertion and Reason are false statements.

Ans. 4
Sol.

183.

Higher conductivity of p-type and n-type semiconductors is due to :

1. Increase in the number of positive holes and mobile electrons respectively  .

2. Increase in the number of electrons in both types of semiconductors.

3. Increase in the number of positive-holes in both types of semiconductors.

4. None of the above.

Ans. 1
Sol.

184.

A: LiCl shows pink colour, when heated in presence of lithium vapours. 

R: LiCl shows Frenkel defect.

1. Both Assertion & Reason are true and the reason is the correct explanation of the assertion.

2. Both Assertion & Reason are true but the reason is not the correct explanation of the assertion.

3. Assertion is true statement but Reason is false.

4. Both Assertion and Reason are false statements.

Ans. 3
Sol.

185.

Consider the following statements about semiconductor.: 

I. Silicon doped with electron-rich impurity is a p-type semiconductor.

II. Silicon doped with an electron-rich impurity is an n-type semiconductor.

III. Delocalised electrons increase the conductivity of doped silicon.

IV. An electron vacancy increases the conductivity of n-type semiconductor.

The correct statement about semiconductors  is - 

1. (I, II)

2. (II, III)

3. (III, IV)

4. (I, IV)

Ans. 2
Sol.

SECTION B

186.

The correct order of increasing bond length of  C-H, C-O, C-C and C=C is-

1.  C-C<C=C<C-O<C-HCC<C=C<CO<CH

2.  C-O<C-H<C-C<C=CCO<CH<CC<C=C

3.  C-H<C-O<C-C<C=CCH<CO<CC<C=C

4.  C-H<C=C<C-O<C-CCH<C=C<CO<CC

Ans. 4
Sol.

187.

The pair that represents chain isomer is :-

1. 

2. 

3. 

4. 

Ans. 3
Sol.

188.

Consider the following resonating structures of HCOOH

The order of stability is-

1. I>II>III>IV

2. IV>I>II>III

3. I>III>II>IV

4. II>I>III>IV

Ans. 1
Sol.

189.

Match the type of mixture of compounds in Column I with the technique of separation/purification given in column II.

Column I

Column II

A. Two solids which have different solubilities in a solvent and which do not undergo a reaction when dissolved in it

1. Steam distillation

B. Liquid that decomposes at its boiling point

2. Fractional distillation

C. Steam volatile liquid

3. Crystallisation

D. Two liquids that have boiling points close to each other

4. Distillation under reduced pressure

Codes

         A          B          C          D      

1.      3          4           1           2

2.      1          2           3           4

3.      1          4           3           2

4.      4          1           3           2

Ans. 1
Sol.

190.

0.24 g of an organic compound containing phosphorous gave 0.66 g of Mg2P2O7 by the usual analysis. The percentage of phosphorous in the compound is-

1. 77%

2. 72%

3. 87%

4. 60 %

Ans. 1
Sol.

191.

In the following reaction,


Product A is-

1. 

2. 

3. 

4. 

Ans. 1
Sol.

192.

What is the end product of the following sequences of operations?

         CaC2H2OCaC2H2OAHg2+Dil.H2SO4Hg2+Dil.H2SO4H2NiH2NiC

1. Methyl alcohol 

2. Acetaldehyde

3. C2H5OHC2H5OH   

4. C2H4C2H4

Ans. 3
Sol.

193.

CH3ClNaCH3ClNaGas A

CH2COONa-CH2COONaCH2COONaCH2COONaElectrolysisKolbe'sElectrolysisKolbe'sGas B

Mg2C3H2OMg2C3H2OGas C

Gas (A+B+C) Cu2Cl2AmmoniacalCu2Cl2Ammoniacal

gas mixture Alk. KMnO4Cold dilAlk. KMnO4Cold dil Gas X

Gas X is-

1. Gas A + Gas B                      2. Only Gas B

3. Gas B + Gas C                      4. Only Gas A

Ans. 4
Sol.

194.

CH3Cl          CH4CH3Cl          CH4

Above conversion can be achieved by:

(1) Zn/H+

(2) LiAlH4

(3) Mg /(ether) then H2O

(4) all of these

Ans. 4
Sol.

195.

(Only one enantiomer is taken)
Which of the following statement is correct about A and B?
(a) A  and B are mixture of diastereomers
(b) A and B are mixture of enantiomers
(c) A and B are optically active
(d) B is racemic mixture

Ans. 1
Sol.

196.

The main air pollutant among the following is : 

1. CO

2. CO2CO2

3. N2N2

4. Sulphur

Ans. 1
Sol.

197.

If a lake is contaminated with DDT, it's highest concentration would be found in :

1. Primary consumer

2. Secondary consumer

3. Tertiary consumer

4. None of the above.

Ans. 3
Sol.

198.

The correct order of the packing efficiency in different types of unit cells is-

1.  Fcc < Bcc < Simple cubic

2.  Fcc > Bcc > Simple cubic

3.  Fcc < Bcc > Simple cubic

4.  Bcc < Fcc > Simple cubic

 

Ans. 2
Sol.

199.

The edge lengths of the unit cells in terms of the radius of spheres constituting fcc, bcc, and simple cubic unit cells are respectively -

1.   22r, 4r3, 2r22r, 4r3, 2r

2.   4r3, 22r, 2r4r3, 22r, 2r

3.   2r, 22r, 4r32r, 22r, 4r3

4.   2r, 4r3, 22r2r, 4r3, 22r

 

Ans. 1
Sol.

200.

(i) Ge is doped with In

(ii) Si is doped with B. 

The semiconductors formed in the above cases are :

1. (i) n-type semiconductor  (ii) p-type semiconductor .

2. (i) p-type semiconductor  (ii) n-type semiconductor .

3. p-type semiconductor in both the cases.

4. n-type semiconductor in both the cases.

Ans. 3
Sol.

PHYSICS

SECTION A

151.

Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Then the change in internal energy of the gas is:

1. 0 

2. 5 J

3. 1 J

4. 3 J

Ans. 1
Sol.
Hint: Work done by the gas in chamber A is zero as chamber B is a vacuum(having zero pressure). Work is only done if there is external pressure.
Step 1: Find work done and heat.


152.

A sample of 0.1 g of water at 1000 C1000 C and normal pressure (1.013 × 105 Nm–2) requires 54 cal of heat energy to convert it into steam at 1000 C1000 C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample is:

1. 104.3 J

2. 208.7 J

3. 42.2 J

4. 84.5 J

Ans. 2
Sol.

According to the first law of thermodynamics
Q=U+WU=Q-WU=54×4.2-PVU=54×4.18-1.013×105167.1×10-6-0.1×10-6U=208.7 JouleΔQ=ΔU+ΔWΔU=ΔQΔWΔU=54×4.2P(ΔV)ΔU=54×4.181.013×105(167.1×1060.1×106)ΔU=208.7 Joule


153.

In a given process dW = 0, dQ < 0, then for the gas:

1.  Temperature -  increases

2.  Volume - decreases

3.  Pressure - decreases

4.  Pressure - increases

Ans. 3
Sol.

From the first law of thermodynamic

dU < 0

Temperature decreases because U = f(T). As the temperature decreases, the pressure also decreases for an isochoric process.


154.

Figure shows the P-V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.

        

(a) Change in internal energy is the same in IV and III cases, but not in I and II.

(b) Change in internal energy is the same in all four cases.

(c) Work done is maximum in case I.

(d) Work done is minimum in case II.

(1) (b, c, d)

(2) (a, d)

(3) (b, c)

(4) (a, c, d)

Ans. 3
Sol.
(3) Hint: The internal energy is a path-dependent variable.
Step 1: Find the change in internal energy.
Change in internal energy for the process A to B,
                                    dUAB=nCvdT=nCv(TB-TA)dUAB=nCvdT=nCv(TBTA)
which depends only on temperatures at A and B.
Step 2: Find the work done.
Work done for A to B, dWAB=dWAB=Area under the P-V curve which is maximum for the path I.

155.

The pressure of a monoatomic gas increases linearly from 4×1054×105 N/m2 to 8×1058×105 N/m2 when its volume increases from 0.2 m3 to 0.5 m3. The work done by the gas is:

1. 2.8×105 J2.8×105 J

2. 1.8×106 J1.8×106 J

3. 1.8×105 J1.8×105 J

4. 1.8×102 J1.8×102 J

Ans. 3
Sol.


Work done by the gas,

W=PdV=area under P-V curve= P1VF-VI+12PF-PI×VF-VI=12VF-VIPF+PI=120.5-08+4×105=1.8×105 JΔW=PdV=area under PV curve= P1(VFVI)+12(PFPI)×(VFVI)=12(VFVI)(PF+PI)=12(0.50)(8+4)×105=1.8×105 J


156.

When a cycle tyre suddenly bursts, the air inside the tyre expands. This process is:
1. isothermal
2. adiabatic
3. isobaric
4. isochoric

Ans. 2
Sol.

Since, in the bursting of the tyre, the gas expands without any exchange of heat, so it is an adiabatic process.


157.

When a system is taken from the state a to state b along the path acb, it is found that a quantity of heat Q = 200 J is absorbed by the system and a work W = 80 J is done by it. Along the path adb, heat absorbed Q = 144 J. The work done along the path adb is:

1. 6 J

2. 12 J

3. 18 J

4. 24 J

Ans. 4
Sol.

158.

Which of the following is true for the molar heat capacity of an ideal gas?

1. It cannot be negative.

2. It has only two values (CP and CV).(CP and CV).

3. It can have any value.

4. It cannot be zero.

Ans. 3
Sol.
Hint: Molar heat capacity of gas can have any value between  to -∞. to ∞.
Step 1: Use the formula of molar heat capacity.
As C=QnTC=ΔQnΔT
where n = number of moles, ΔQ = heat absorbed by the gas, and ΔT = rise in temperature of the gas.
Step 2: Find the molar heat capacity for the isothermal and adiabatic process.
Cisothermal= (As T=0)Cadiabatic=0 (As Q=0)Cisothermal= (As ΔT=0)Cadiabatic=0 (As ΔQ=0)
Hence, the molar heat capacity can have any value.


159.

An ideal gas heat engine operates in a Carnot cycle between 227ºC and 127ºC. It absorbs 6 × 104 cals of heat at higher temperatures. Amount of heat converted to work is:-

(1) 4.8 × 104 cals

(2) 2.4 × 104 cals

(3) 1.2 × 104 cals

(4) 6 × 104 cals

Ans. 3
Sol.

η=WQ=1T2 T1W=Q(1T2 T1)=1.2×104calsη=WQ=1T2 T1W=Q(1T2 T1)=1.2×104cals


160.

During an experiment an ideal gas is found to obey an additional law VP2 = constant. The gas is initially at temp. T and volume V. What will be the temperature of the gas when it expands to a volume 2V?

1. T'=4TT'=4T

2. T'=2TT'=2T

3. T'=5TT'=5T

4. T'=6TT'=6T

Ans. 2
Sol.

According to the given problem:

VP2 = constant

From the gas law,

PV = nRT

(kV)V=nRTV=(nRK)TV1V2=(T1T2) i.e., V2V=TT'T'=2T(kV)V=nRTV=(nRK)TV1V2=(T1T2) i.e., V2V=TT'T'=2T


161.

Compound A (C4H8C4H8) is treated with H2O/H2SO4 gives C4H10OC4H10O, an optically inactive compound. The structure of A is-

1. CH3CH2CH=CH2CH3CH2CH=CH2                       

2. CH3CH=CHCH3CH3CH=CHCH3

3. CH32C=CH2(CH3)2C=CH2                               

4.

Ans. 3
Sol.

Hint: The number of moles is conserved.

Step 1: Find the initial no. of moles.

PV=nRT

2n=2P0V0RT02n=2P0V0RT0

Step 2: Find the final no. of moles.

nf=P'V0RT+P'V0RT0nf=P'V0RT+P'V0RT0

Step 3: Conserve the no. of moles.

ni=nf2P0V0RT0=P'V0RT0+P'V0RT2P0T0=P'T+T0TT0P'= 2TP0T+T0ni=nf2P0V0RT0=P'V0RT0+P'V0RT2P0T0=P'(T+T0)TT0P'= 2TP0T+T0


162.

The ratio of the average translatory kinetic energy of He gas molecules to O2O2 gas molecules is:

1. 25212521

2. 21252125

3. 3232

4. 1

Ans. 4
Sol.
 
Hint: Average K.E. of molecules = 3/2 KBT
 
Step 1: Find the average kinetic energy of a molecule
K=32kTK=32kT
 
Step 2: Find the average kinetic energy of molecules
K=32RTK=32RT

163.

When an ideal gas is compressed adiabatically, its temperature rises: the molecules on average have more kinetic energy than before. The kinetic energy increases,

(1) because of collisions with moving parts of the wall only.

(2) because of collisions with the entire wall.

(3) because the molecules get accelerated in their motion inside the volume.

(4) because of the redistribution of energy amongst the molecules.

Ans. 1
Sol.

Hint: The kinetic energy of a gas depends on the collisions between the molecules. When the gas is compressed adiabatically, the total work done on the gas increases its internal energy which in turn increases the kinetic energy of gas molecules and hence, the collisions between molecules also increase.


164.

To find out degree of freedom, the correct expression is :

1. f=2γ-1f=2γ1

2. f=γ+12f=γ+12

3. f=2γ+1f=2γ+1

4. f=1γ+1f=1γ+1

Ans. 1
Sol. xmlns&#x2235;&#xA0;&#x3B3;=1+2f&#x21D2;2f=&#x3B3;-1&#x21D2;f=2&#x3B3;-1" role="presentation" style="font-size: 113%; min-width: 10.755em; position: relative;"> γ=1+2f2f=γ-1f=2γ-1 γ=1+2f2f=γ1f=2γ1

165.

The figure shows a process for a gas in which pressure (P) and volume (V) of gas change. If C1C1 and C2C2 are the molar heat capacities of gas during process AB and BC respectively, then

1. C1=C2C1=C2

2. C1>C2C1>C2

3. C1<C2C1<C2

4. C1C2C1C2

Ans. 3
Sol.
Hint: Use Cp-Cv = R.CpCv = R.
 
Step 1: Identify the processes
AR - Isochoric
BC - Isobaric
 
Step 2: Find the molar specific heats C1 and C2 
C1=CvC2=CpCp-Cv=RCp = Cv+RCp > CvHence, C2 > C1C1=CvC2=CpCpCv=RCp = Cv+RCp > CvHence, C2 > C1

166.

The root mean square velocity of the gas molecules is 300 m/s. What will be the root mean square speed of the molecules if the atomic weight is doubled and absolute temperature is halved?

1. 300 m/s

2. 150 m/s

3. 600 m/s

4. 75 m/s

Ans. 2
Sol.
Hint: Recall the formula for rms speed.

Step 1: Use formula for rms speed.

vrms=3RTMvrmsTMv2v1=T2T1×M1M2=12v2=v12=150 m/svrms=3RTMvrmsTMv2v1=T2T1×M1M2=12v2=v12=150 m/s


167.

At room temperature, the r.m.s. speed of the molecules of certain diatomic gas is found to be 1930 m/s. The gas is :

1. H2H2                               

2. F2F2

3. O2O2                               

4. Cl2Cl2

Ans. 1
Sol.

1.   vrms=3RTM        M=3RTvrms2       M=3×8.3×30019202        =2×10-3 kg  =2 gm         Gas is hydrogen.1.   vrms=3RTM        M=3RTv2rms       M=3×8.3×300(1920)2        =2×103 kg  =2 gm         Gas is hydrogen.


168.

When the gas in an open container is heated, the mean free path:

1. Increases

2. Decreases

3. Remains the same

4. Any of the above depending on the molar mass

Ans. 1
Sol.
Hint: At constant volume and pressure, T1n1nAs temperature increases moles of gas decrease.
Step 1: Apply the formula of the mean free path.
As
λ=12πnvd2Herenv=Number densityStep 2: Use the formula of number density.nvMolesVolumeas volume is constant so, nv is decreasing and λ will increase.λ=12πnvd2Herenv=Number densityStep 2: Use the formula of number density.nvMolesVolumeas volume is constant so, nv is decreasing and λ will increase.

169.

Identify the correct definition.

1. If after every certain interval of time, a particle repeats its motion then motion is called periodic motion.

2. To and fro motion of a particle is called oscillatory motion.

3. Oscillatory motion described in terms of single sine and cosine functions is called simple harmonic motion.

4. All of the above

Ans. 4
Sol.

Hint: Recall the definitions of different types of motions. Periodic motion repeats its motion after every certain interval of time. Simple harmonic motion can be described in terms of single sine or cosine functions.


170.

Displacement versus time curve for a particle executing SHM is shown in the figure. Choose the correct statement/s.

1. Phase of the oscillator is the same at t =0 s and t = 2s

2. Phase of the oscillator is the same at t =2 s and t=6s

3. Phase of the oscillator is the same at t = 1s and t=7 s

4. The phase of the oscillator is the same at t=1 s and t=5 s

(1) 1, 2 and 4

(2) 1 and 3

(3) 2 and 4

(4) 3 and 4

Ans. 3
Sol.
Hint: The phase of the oscillator will be the same after each time period.
Step 1: Find the time period of SHM.
             T=4s
Step 2: Assign a phase for each second.

The phase is same at t = 2 sec & 6 sec and t = 1 sec & 5 sec.

171.

A particle is executing linear S.H.M. between x = A. The time taken to go from 0 to A/2 is T1T1 and to go from A/2 to A is T2T2, then

1.  T1 < T2T1 < T2

2.  T1 > T2T1 > T2

3.  T1 = T2T1 = T2

4.  T1 = 2T2T1 = 2T2

Ans. 1
Sol.

Let the time period of the pendulum = T

Then, time to go from 0 to A = T4T4/

Time to go from 0 to A/2:-

Displacement = A2=A Sinωt1A2=A Sinωt1

Sinωt1=12=Sin π6/ωt1=π62πT×t1=π6t1=T12Sinωt1=12=Sin π6/ωt1=π62πT×t1=π6t1=T12

Time to go from A/2 to A:-

t2=T4-T12=T6T1=T12 & T2=T6T1<T2t2=T4T12=T6T1=T12 & T2=T6T1<T2


172.

The displacement of a particle along the x-axis is given by x=asin2 ωtx=asin2 ωt. The motion of the particle corresponds to:

(1) simple harmonic motion of frequency ω/π.ω/π.

(2) simple harmonic motion of frequency 3ω/2π.3ω/2π.

(3) non-simple harmonic motion.

(4) simple harmonic motion of frequency ω/2π.ω/2π.

Ans. 1
Sol.

x=asin2ωtusing, cos2ωt =12sin2ωtsin2ωt =1cos2ωt2So, x =a2acos2ωt2x=asin2ωtusing, cos2ωt =12sin2ωtsin2ωt =1cos2ωt2So, x =a2acos2ωt2

Reference: NCERT, Class-XI, Part-2 (Oscillations)
Example 14.3: Which of the following functions of time represent (a) simple harmonic motion and (b) periodic but not simple harmonic? Give the period for each case.
(1) sinω tcosω tsinω tcosω t
(2) sin2ω tsin2ω t

Note: sin2ω tsin2ω t is corresponds to simple harmonic motion about x=a/2.x=a/2.

Trying to determine if x=asin2ωtx=asin2ωt represents SHM by: 
d2xdt2+ω2x=0d2xdt2+ω2x=0 will be incorrect. 
The reason is that the origin and point of equilibrium are not coinciding. In the equation, x=asin2ωtx=asin2ωt , the point of equilibrium is (a2 ,0a2 ,0) and origin is (0,0). Therefore, the equation of SHM d2xdt2+ω2x=0d2xdt2+ω2x=0 will not work here. If you shift origin (0,0) to  (a2 ,0a2 ,0) then it will work.

How to  justify that  x=asin2ωtx=asin2ωt  is SHM about x=a2x=a2 ? 
Given,  x=asin2ωtx=asin2ωt 
x=a2(1cos2ωt)x=a2(1cos ωt)x=a2(1cos2ωt)x=a2(1cos ωt)
where  ω=2ωω=2ω
 Now    dxdt=aω2sinωtand  d2xdt2=aω22cosωtNow    dxdt=aω2sinωtand  d2xdt2=aω22cosωt
d2xdt2=aω22[12sin2ωt2]d2xdt2=aω22aω2sin2ωtd2xdt2=ω2[a2asin2ωt]d2xdt2=ω2[a2x]d2xdt2+ω2(xa2)=0d2xdt2=aω22[12sin2ωt2]d2xdt2=aω22aω2sin2ωtd2xdt2=ω2[a2asin2ωt]d2xdt2=ω2[a2x]d2xdt2+ω2(xa2)=0
Hence,  x=asin2ωtx=asin2ωt is SHM of frequency ωπωπ about x=a2x=a2

 


173.

Equation of simple harmonic motion is x = asinωωt, then for which value of x, kinetic energy is equal to the potential energy?

1.  x = ± ax = ± a

2.  x = ± a2x = ± a2

3.  x = ± a2x = ± a2

4.  x = ± 3a2x = ± 3a2

Ans. 3
Sol.
Hint: Kinetic energy is maximum at the mean position whereas potential energy is maximum at the extreme position.
Step 1: Kinetic energy and potential energy respectively are

K=12mw2(A2-x2)Step 2: Potential energy, U=12kx2Step 3:As,K=U12mω2(A2-x2)=12kx2ω=kmmω2=kA2-x2=x2x=±A2.K=12mw2(A2x2)Step 2: Potential energy, U=12kx2Step 3:As,K=U12mω2(A2x2)=12kx2ω=kmmω2=kA2x2=x2x=±A2.


174.

The graph between the velocity (v) of a particle executing S.H.M. and its displacement (x) is shown in the figure. The time period of oscillation is:

                    

1.  αβαβ

2.  2παβ2παβ

3.  2π(βα)2π(βα)

4.  2π(αβ)2π(αβ)

Ans. 4
Sol.
Hint: vmax=ωAvmax=ωA
Step 1: Find the maximum velocity and displacement.
vmax=Aω=βxmax=A=αAωA=βαω=βαvmax=Aω=βxmax=A=αAωA=βαω=βα

Step 2: Find the time period.

T=2πω=2παβT=2πω=2παβ


175.

A pendulum has time period T. If it is taken on to another planet having acceleration due to gravity half and mass 9 times that of the earth then its time period on the other planet will be

(1) TT

(2) T

(3) T1/3T1/3

(4) 22T

Ans. 4
Sol.

 

 T=2πlgT1gTpTe=gegp=21Tp=2TT=2πlgT1gTpTe=gegp=21Tp=2T


#53947

176.

The time period of the spring-mass system depends upon

1.  the gravity of earth

2.  the mass of block

3.  spring constant

4.  Both 2 & 3

Ans. 4
Sol.

Hint - T = 2πmK2πmK

Option(4) is correct.

 


177.

A mass of 30 g is attached with two springs having spring constant 100 N/m and 200 N/m and other ends of springs are attached to rigid walls as shown in the given figure. The angular frequency of oscillation is

                        

1.  1002π rad/s1002π rad/s

2.  100π rad/s100π rad/s

3.  100 rad/s

4.  200ππ rad/s

Ans. 3
Sol.
Class XI, Part 2, pg 352
Hint: Find equivalent spring constant
Step 1: Check if springs are connected in series or parallel
keq=k1+k2=100+200=300 N/mkeq=k1+k2=100+200=300 N/m
Step 2: Find angular frequency
ω=keqm=30030×10-3=100 rad/sω=keqm=30030×103=100 rad/s

178.

In damped oscillations, the damping force is directly proportional to the speed of the oscillator. If amplitude becomes half of its maximum value in 1 sec, then after 2 sec, the amplitude will be: (Initial amplitude = A0A0)

1. 14A014A0

2. 12A012A0

3. A0A0

4. 3A023A02

Ans. 1
Sol.
Hint: Use the formula for the amplitude of damped oscillations.

Step 1: Use the equation for the amplitude. 

A=A0e-bt2mAfter t=1sA02=A0e-b2me-b2m=12   ........(1)A=A0ebt2mAfter t=1sA02=A0eb2meb2m=12   ........(1)

Step 2: Apply equation at t=2s.

A=A0e-2b2m=A0(e-b2m2=A0122=14A0A=A0e2b2m=A0((eb2m)2=A0(12)2=14A0


179.

Which one of the following statements is true?

1. Both light and sound waves in air are transverse

2. The sound waves in the air are longitudinal while the light waves are transverse

3. Both light and sound waves in air are longitudinal

4. Both light and sound waves can travel in a vacuum

Ans. 2
Sol.

In a longitudinal wave, the particles of the medium oscillate about their mean or equilibrium position along the direction of propagation of the wave itself. Sound waves are longitudinal in nature. In transverse waves, the particles of the medium oscillate about their mean or equilibrium position at right angles to the direction of propagation of the wave itself. Light waves being electromagnetic are transverse waves.


180.

Eleven tuning forks are arranged in increasing order of frequency in such a way that any two consecutive tunning forks produces 4 beats per second. The highest frequency is twice that of the lowest. The frequency is twice that of the lowest. The highest and the lowest frequencies (in Hz) are respectively

1.  100 and 50

2.  44 and 22

3.  80 and 40

4.  72 and 30

Ans. 3
Sol.

Let lowest frequency = f

Then, highest frequency=f+(11-1)××4=f+40

Now, f+40=2ff=40

Highest frequency=80 Hz

Lowest frequency=40 Hz


181.

A transverse wave moves from a medium A to a medium B. In medium A, the velocity of the transverse wave is 500 ms11 and the wavelength is 5 m. The frequency and the wavelength of the wave in medium B when its velocity is 600 ms11, respectively are:
1. 120 Hz and 5 m
2. 100 Hz and 5 m
3. 120 Hz and 6 m
4. 100 Hz and 6 m

Ans. 4
Sol.
Frequency of the wave,
f=v1λ1=v2λ2f=5005=100 HzFrequency of the wave remians the same.f=600λ2λ2=600100=6mf=v1λ1=v2λ2f=5005=100 HzFrequency of the wave remians the same.f=600λ2λ2=600100=6m

182.

Two identical piano wires kept under the same tension T have a fundamental frequency of 600 Hz. The fractional increase in the tension of one of the wires which will lead to the occurrence of 6 beats/s when both the wires oscillate together would be :

(1) 0.02                                           

(2) 0.03

(3) 0.04                                             

(4) 0.01 

Ans. 1
Sol.

Hint: The beat frequency is given by the difference of frequencies of the piano wires.

Step 1: Find the relation between the tension and the frequency.

ν=nv2LνvTStep 2: Find the relative change in tension.νν = 12TTTT = 2 ×νν = 2 ×6600 = 0.02ν=nv2LνvTStep 2: Find the relative change in tension.Δνν = 12ΔTTΔTT = 2 ×Δνν = 2 ×6600 = 0.02

 


183.

A source of sound with frequency n = 2000 Hz moves along a line at right angles to the wall with a velocity vs=0.33vs=0.33 m/s. Two stationary detectors D1 and D2D1 and D2 are placed on the path of the source as shown in the figure. The velocity of the sound in air is v = 330 m/s. Then:

1. Only D1D1 records beats

2. Only D2D2 records beats

3. Both D1 and D2D1 and D2 record beats

4. Neither D1 nor D2D1 nor D2 record beats

Ans. 1
Sol. <
 
Hint: Recall Doppler's effect.
 
Step 1: Find apparent frequency for the wall of the sound source.
 
f0=nv+v0v-vs=2000330330-0.33f0=n[v+v0vvs]=2000[3303300.33]
 
Step 2: Find apparent frequency for detectors of the sound source S.
 
fD1=2000330+0330+0.33fD2=2000330+0330-0.33fD1=2000[330+0330+0.33]fD2=2000[330+03300.33]
 
Step 3: Find apparent frequency of reflected sound from wall heard by detectors D1 & D2.D1 & D2.
As there is no relative motion between the detectors and the wall, the detectors will hear the sam frequency of the reflected sound.
Step 4: Check for beats.
 
D1 will hear the beats as two different frequencies are reaching to it.D2 will hear no beats as it is receiving the same frequency.D1 will hear the beats as two different frequencies are reaching to it.D2 will hear no beats as it is receiving the same frequency.

184.

Mark out the correct options.

1. The energy of any small part of a string remains constant in a travelling wave.

2. The energy of any small part of a string remains constant in a standing wave.

3. The energies of all the small parts of equal length are equal in a travelling wave.

4. The energies of all the small parts of equal length are equal in a standing wave.

Ans. 3
Sol.

 Use I1r2 Use I1r2


185.

Two vibrating tuning forks produce progressive waves given by Y1 = 4sin500πt and Y= 2sin506πt. The number of beats produced per minute is:

(1) 3

(2) 360

(3) 180

(4) 60

Ans. 3
Sol.

 Number of beats per second =n2n1=3(n1=250,n2=253) Number of beats produced per minute =3×60=180 Number of beats per second =n2n1=3(n1=250,n2=253) Number of beats produced per minute =3×60=180


SECTION B

186.

For a gas, RCV=0.67RCV=0.67.  This gas is made up of molecules which are:                                                                   [JIPMER 2001, 02]

1. Diatomic           

2. Mixture of diatomic and polyatomic molecules

3. Monoatomic     

4. Polyatomic

Ans. 3
Sol.

(c)

CV=R0.67=1.5 R=32RCV=R0.67=1.5 R=32R

This is the value of CVCV for monoatomic gases.


187.

The carbocation among the following doesn't stabilized by resonance is-

1. 

2. 

3. 

4. 

Ans. 4
Sol.

(d)

pV=μRT=mMRT pVT1M     (M = Molecular mass)From graph  pVTA<pVTB<pVTC                     MA>MB>MCpV=μRT=mMRT pVT1M     (M = Molecular mass)From graph  (pVT)A<(pVT)B<(pVT)C                     MA>MB>MC


188.

If CpCp and CvCv denote the specific heats (per unit mass) of an ideal gas of molecular weight M

(1) Cp-Cv=RM2CpCv=RM2                             

(2) Cp-Cv=RCpCv=R

(3) CpCv=RMCpCv=RM                               

(4) CpCv=MRCpCv=MR

Ans. 3
Sol.

             CpCv=R  MCpMCv=R              CpCv=RM             CpCv=R  MCpMCv=R              CpCv=RM



189.

At 0°C the density of a fixed mass of a gas divided by pressure is x. At 100°C, the ratio will be:

1. xx                                   

2. 273373x273373x

3. 373273x373273x                           

4. 100273x100273x

Ans. 2
Sol.

2.   PV=μRT=mMRT mVP =densityP=MRT       (densityP)At 0°C=MR(273)=x                          ..........(i)       (densityP)At 100°C=MR(373)                             ..........(ii)        (densityP)At 100°C=273 373x2.   PV=μRT=mMRT mVP =densityP=MRT       (densityP)At 0°C=MR(273)=x                          ..........(i)       (densityP)At 100°C=MR(373)                             ..........(ii)        (densityP)At 100°C=273 373x


190.

A gas undergoes the change in its thermodynamic state from A to B via two different paths as shown in the given pressure (P) versus volume (V) graph, then:

1. the temperature of the gas decreases in path 1 from A to B.

2. the heat absorbed by the gas in path 1 is greater than in path 2.

3. the heat absorbed by the gas in path 2 is greater than in path 1.

4. the change in internal energy in path 1 is greater than in path 2.

Ans. 2
Sol.
Hint: Find the work done for paths.
Step 1: Find temperature change in path 1.
Step 2: Compare the work done in both paths.
W1>W2W1>W2 
Step 3: Find the change in internal energy for both paths.
U=Q-WQ1-W1=Q2-W2Q1>Q2ΔU=ΔQWΔQ1W1=ΔQ2W2ΔQ1>ΔQ2

191.

In the following figures, four curves A, B, C and D are shown. The curves are:

                      

1. isothermal for A and D while adiabatic for B and C.

2. adiabatic for A and C while isothermal for B and D.

3. isothermal for A and B while adiabatic for C and D.

4. isothermal for A and C while adiabatic for B and D.

Ans. 4
Sol.

Step 1: Find the slope for the curves.

PVx=constantdP(Vx)+xVx-1(dV)P=0dPdV=-xPVIsothermal: x=1Slope=PVAdiabatic: x=ySlope=yPVPVx=constantdP(Vx)+xVx1(dV)P=0dPdV=xPVIsothermal: x=1|Slope|=PVAdiabatic: x=y|Slope|=yPV


192.

For one complete cycle of a thermodynamic process on a gas as shown in the P-V diagram, which of following is correct ?

1. ΔEint=0,Q<0ΔEint=0,Q<0

2. ΔEint=0,Q>0ΔEint=0,Q>0

3. ΔEint>0,Q<0ΔEint>0,Q<0

4. ΔEint<0,Q>​ 0ΔEint<0,Q>0

Ans. 4
Sol.

Eint=0W=-veQ=Eint+WQ<0ΔEint=0W=veQ=ΔEint+WQ<0


193.

The temperature inside a refrigerator (reversible process) is t2oC and the room temperature is t1oC. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be:

1.  t1t1-t2t1t1t2

2.  t1+273t1-t2t1+273t1t2

3.  t2+273t1+t2t2+273t1+t2

4.  t1+t2t1+273t1+t2t1+273

Ans. 2
Sol.

For a refrigerator, we know that:
Q1W=Q1Q1-Q2=T1T1-T2Q1W=Q1Q1Q2=T1T1T2 
where, 
Q1 = amount of heat delivered to the room 
W = electrical energy consumed 
T1 =room temperature = t1 + 273 
T2 =temperature of sink= t2 + 273 
 Q11=t1+273t1+273-t2+273Q1=t1+273t1-t2Q11=t1+273t1+273(t2+273)Q1=t1+273t1t2


194.

The velocity-time diagram of a harmonic oscillator is shown in the adjoining figure. The frequency of oscillation is

(1) 25 Hz

(2) 50 Hz

(3) 12.25 Hz

(4) 33.3 Hz

Ans. 1
Sol.

 f=1T=10.04=25 Hzf=1T=10.04=25 Hz


195.

Kinetic energy of a particle executing simple
harmonic motion in straight line is pv2 and
potential energy is qx2, where v is speed at
distance x from the mean position. Its time
period is given by the expression

1. 2πqp2πqp

2. 2πpq2πpq

3. 2πqp+q2πqp+q

4. 2πpp+q2πpp+q

Ans. 2
Sol.

Hint: Recall the general equation for KE and PE

Step 1: 

K=12mv2=pv2p=m2U=122x2=qx2q=22qp=ω2ω=qpK=12mv2=pv2p=m2U=12mω2x2=qx2q=mω22qp=ω2ω=qp

Step 2: Find time period

T=2πω=2πpqT=2πω=2πpq


196.

On a smooth inclined plane, a body of mass M is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has force constant K, the period of oscillation of the body (assuming the springs as massless) is

                

(1) 2πM2K1/22π(M2K)1/2

(2) 2π2MK1/22π(2MK)1/2

(3) 2πMg sinθ2K2πMg sinθ2K

(4) 2π2MgK1/22π(2MgK)1/2

Ans. 1
Sol.

(a)

Let the initially, elongation in the spring=yMgsinθ=2kyDisplace the block by x;Fnet=Mgsinθ-2ky-2kx=-2kxT=2πM2kLet the initially, elongation in the spring=yMgsinθ=2kyDisplace the block by x;Fnet=Mgsinθ2ky2kx=2kxT=2πM2k


197.

The displacement x of a particle varies with time t as x = A sin2πTt + π3x = A sin(2πTt + π3). Time taken by particle to reach from x = A2x = A2 to x = -A2x = A2 is-

1.  T2T2

2.  T3T3

3.  T12T12

4.  T6T6

Ans. 4
Sol.
Hint: Use a circle diagram to find the time taken by particle to reach from x = A2x = A2 to x = -A2x = A2.
Step 1: Draw the circle diagram.

Step 2: Find the angle with the help of diagram.

As,sinθ=A/2A=12θ=30°θ=π3Step 3: Find the time taken,t=θωHere,ω=2πTt=T6As,sinθ=A/2A=12θ=30°Δθ=π3Step 3: Find the time taken,t=ΔθωHere,ω=2πTt=T6


198.

In an experiment with a sonometer, a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. The tension of the string remaining constant, the frequency of the second tuning fork is -

1. 163.84 Hz

2. 400 Hz

3. 320 Hz

4. 204.8 Hz

Ans. 2
Sol.
Hint: In the case of the sonometer both the ends of the string are fixed.
Step 1: At resonance, the frequency of the tuning fork is the same as the frequency of the sonometer.
The frequency of the sonometer is given as
f=12lTμf1lStep 2: Calculate the value of f2 using the above relation.Hence,f2f1=l1l2f2=256×2516=400 Hz.f=12lTμf1lStep 2: Calculate the value of f2 using the above relation.Hence,f2f1=l1l2f2=256×2516=400 Hz.

199.

The percentage increase in the speed of transverse waves produced in a stretched string if the tension is increased by 4%, will be:

1. 1%

2. 2%

3. 3%

4. 4%

Ans. 2
Sol.

Hint: Speed of transverse wave = TμTμ

Step 1: Find the %age change in the speed.

v =  TμTμ

(Δvv×100)=12(ΔTT×100)=12×4=2%(Δvv×100)=12(ΔTT×100)=12×4=2%


200.

 

A source of sound S emitting waves of frequency 100 Hz and an observer O is located at some distance from each other. The source is moving with a speed of 19.4 ms-1 at an angle of 600600 with the source-observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 ms-1), is: 

1. 100 Hz

2. 103 Hz

3. 106 Hz

4. 97 Hz

Ans. 2
Sol.


 
The apparent frequency heard by the observer,
fo=fsvv-vscos60°=100330330-19.4×12100330330-9.7=100330320.3fo=fs[vvvscos60°]=100[33033019.4×12]100[3303309.7]=100[330320.3] 
 
=103.02 Hz